Calculate the summation of double continued fractions

Question

A few month ago, my brother had given me this question: \begin{equation} \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{\cdots+\frac{1}{2005}} } } }+\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{\cdots+\frac{1}{2005}} } } } \end{equation} He said to me, "this is called continued fraction and the answer equals 1." Since then, I learned from many websites about CF but I failed to prove the summation of those CFs equals 1. I only see this pattern: \begin{equation} \cfrac{1}{2 + \frac{1}{3} }=\frac{\color{green}3}{\color{red}7}\text{ and }\cfrac{1}{1 + \cfrac{1}{2+\cfrac{1}{3}} }=\frac{\color{red}7}{10}=\frac{\color{red}7}{\color{green}3+\color{red}7}\text{, etc.} \end{equation} Let's say the first CF equals $\frac{a}{b}$, then the second CF will equal $\frac{b}{a+b}$. Hence \begin{equation} \frac{a}{b}+\frac{b}{a+b}=\frac{a^2+ab+b^2}{ab+b^2}=1+\frac{a^2}{ab+b^2}>1 \end{equation} Is this correct? Did my big brother trick me all this time? Someone here please help me, preferably someone with a doctorate degree in math or a college math professor so I can be sure that I'm right and can argue with him. I want to win this time because he always shows off his smartness to me. BTW, I'm just an 8th grade student, so please be nice to me. Thank you. :)

Answer 1

First, I would like to reiterate that:

1. Your brother's claim is incorrect
2. The proof you gave that your brother is wrong is quite correct
3. It is very similar to what I would have written myself

Just for completeness, here is what I would have said: Let $x$ be the left-hand term in your brother's expression. Then the right-hand term is equal to $\frac1{1+x}$. Your brother then claims that $$x + \frac1{1+x} = 1.$$ Simplifying the left-hand side we get $$\frac{x^2+x+1}{x+1} = 1$$ which can happen only if $$x^2+x+1 = x+1$$ so $x^2 = 0$ and $x=0$. But clearly $x>0$, so your brother's claim is mistaken. This is exactly what you said, except you had $\frac ab$ in place of $x$.

You might also observe that this argument works even if your brother carries the continued fractions on past $2005$ to infinity.

The rest of this note is about continued fractions a little more generally.

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Instead of the clumsy notation $$z=a_0+ \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3+ \cfrac{1}{\cdots+\frac{1}{a_n}} } } }$$ let's agree to write $z=[a_0; a_1, a_2,\ldots a_n]$. Let's consider what happens when we stop early in expanding $z$ and write $$\begin{align}z_0 &= a_0\\ z_1 & = [a_0; a_1] = a_0 + \frac1{a_1}\\ z_2& = [a_0; a_1, a_2] = a_0+\cfrac{1}{a_1+\cfrac1{a_2}}\\z_3&=[a_0; a_1,a_2,a_3]\\& \;\vdots\\ z = z_n & = [a_0; a_1,\ldots, a_n]\end{align}$$

These $z_i$ are called the convergents of $z$. When $i$ is even we have the even convergents and when $i$ is odd we have the odd convergents.

Here is the single most important thing to know about continued fractions:

$$z_0\lt z_2 \lt z_4\lt\ldots < z < \ldots < z_5 < z_3 < z_1 \tag{$\star$}$$

The even convergents of $z$ form an increasing sequence that approaches the value $z$ from below, while the odd convergents form a decreasing sequence that approaches $z$ from above.

Let's take $x = [0; 2, 3, 4, \ldots, 2005]$ as in your brother's example. Taking just the first two convergents we have $0 < x < [0;2] = \frac12$, and taking the next two we have $$0 + \cfrac1{2+\cfrac13} < x < \cfrac1{2+\cfrac1{3+\cfrac 14}}$$ so $$\frac37 < x < \frac{13}{30}\\0.4286\ldots < x < 0.4333\ldots$$ which narrows down the possible value of $x$ rather dramatically. Even without using the computer to calculate the exact value, we now know it is not much more than $\frac37$.

Similarly, let $y$ be the right-hand term of your brother's expression, $[0;1,2,3\ldots, 2005]$. Then we have $$[0] < y < [0;1] = 1\\ [0;1,2] = \frac23 = 0.6666\ldots < y < [0;1,2,3] = \frac7{10} = 0.7.$$

Already this is enough to prove that your brother's claim is mistaken, because we have $$1 < \frac{23}{21} = \frac37 + \frac 23 < x + y .$$

The book Continued Fractions by A. Ya. Khinchin is short, and the first part is not too hard; you might take a look at it. Theorem 4 on page 6 is the important fact $(\star)$ that I mentioned above.